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Inner product with maximally entangled state

  1. Dec 14, 2011 #1
    Consider the maximum inner product of a density operator with a maximally entangled state. (So, given a density operator, we're maximizing over all maximally entangled states.)

    I'm pretty sure the minimum value (a lower bound on the maximum) for this is 1/n2, using the density operator 1/n2 identity. How can I prove that is the minimum?

    Thanks!
    Reuben
     
  2. jcsd
  3. Dec 15, 2011 #2

    tom.stoer

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    Science Advisor

    can you please write down the density operator you are talking about; and what you mean by 'maximum inner product of a density operator with a maximally entangled state'.
     
  4. Dec 15, 2011 #3
    Sorry I realize that was a little incomplete. Let [itex]\mathcal{X}[/itex] and [itex]\mathcal{Y}[/itex] be complex Euclidean spaces with dim([itex]\mathcal{X}[/itex])=dim([itex]\mathcal{Y}[/itex])=n.

    Define

    [itex]
    M(\rho) = \max\left\{<u u^{\ast},\rho>\,:\,u\in\mathcal{X}\otimes\mathcal{Y}\;\text{is maximally entangled}
    \right\}
    [/itex]

    for any density operator [itex]\rho[/itex] on [itex]\mathcal{X}\otimes\mathcal{Y}[/itex].

    For [itex]\rho=\frac{1}{n^2}\text{I}[/itex] we get [itex]M=\frac{1}{n^2}[/itex]

    How can I show that that is the minimum value of [itex]M[/itex] over all possible density operators [itex]\rho[/itex]?
     
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